Quotient group by two isomorphic groups

Question

In the processes of studying some questions I suddenly realized something basic but weird. We have $\mathbb{Z}\cong 2\mathbb{Z}$, but $\mathbb{Z}/\mathbb{Z}\ncong \mathbb{Z}/2\mathbb{Z}$. It looks like isomorphism doesn't mean exactly the same, even in algebra problems. But I remember many tricks are based on that we regard isomorphic groups the same group (e.g. we may consider $G$ is a subgroup of $\text{Aut}(G)$), how can I check if those tricks are correct (or is it necessary to check)?

Answer 1

Short answer

Isomorphism is not equality and you can't expect it to solve all your problems magically.

Long answer

As I said in the comments:

"Basically a quotient depends not only on the isomorphism type of the "numerator" and of the "denominator", but also on the way that the "denominator" is embedded in the "numerator"."

More in general when there is a construction involving many objects it's possible that this construction depends not only on the isomorphism type of the objects but also on the interactions between these object.

Trivial example:

Let's say that two finite sets are isomorphic if they have the same number of elements. Let's consider the sets

$$A_1=\{1\}$$ $$A_2=\{2\}$$ $$B_1=\{1,2\}$$ $$B_2=\{3,4\}$$

Clearly $A_1\cong A_2$ and $B_1 \cong B_2$ but $A_1\cup B_1$ isn't isomorphic to $A_2 \cup B_2$ (because the interaction between $A_1$ and $B_1$ is different from the one between $A_2$ and $B_2$).

For some costructions the interaction doesn't matter like the cartesian product and the disjoint union: the deeper motivation as someone noticed in the comment is that these are what's called universal constructions in category theory (but at your level you don't need to know category theory).